Problem: $\sum\limits_{n=0}^{\infty }{{\left( -1 \right)}^{n}}\frac{{{x}^{4n+2}}}{\left( 2n \right)!}~~$ is the Maclaurin series for which function? Choose 1 answer: Choose 1 answer: (Choice A) A $\sin(x)$ (Choice B) B $\sin(x^2)$ (Choice C) C $x\sin x$ (Choice D) D $\cos x^2$ (Choice E) E $x^2\cos x$ (Choice F) F $x^2\cos (x^2)$
Explanation: We need to write out the first several terms of the series in order to see the important properties. $\sum\limits_{n=0}^{\infty }{{\left( -1 \right)}^{n}}\frac{{{x}^{4n+2}}}{\left( 2n \right)!}~=~x^2-\dfrac{x^6}{2!}+\dfrac{x^{10}}{4!}-\dfrac{x^{14}}{6!}+...$ The given series is alternating $\big($ because of the $(-1)^n\big)$ and has denominators that are all even factorials. This suggests that the series has to do with the function $~\cos x$. However, we also note that the powers of $x$ go up by 4 in each term $-$ not 2 as would be expected. Thus we see this as a function of $~\cos\big(x^2\big)\,$, not just $~\cos x\,$. $ \cos~ x=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{2n}}}{\left( 2n \right)!}+...$ Hence, $ \cos~ (x^2)=1-\frac{{{(x^2)}^{2}}}{2!}+\frac{{{(x^2)}^{4}}}{4!}-...+{{\left( -1 \right)}^{n}}\frac{{{(x^2)}^{2n}}}{\left( 2n \right)!}+...$ When we multiply through by $x^2\,$, we obtain the following: $ x^2 \cos (x^2)=x^2-\frac{{{x}^{6}}}{2!}+\frac{{{x}^{10}}}{4!}-...+{{\left( -1 \right)}^{n}}\frac{{{x}^{4n+2}}}{\left( 2n \right)!}+...$